Section 15.1 Double Integrals over Rectangles
Ruipeng Shen
April 2
1
Volumes and Double Integrals
We consider a function f of two variables defined on a closed rectangle
R = [a, b] × [c, d] = {(x, y) ∈ R2 : a ≤ x ≤ b, c ≤ y ≤ d}
and we suppose that f (x, y) ≥ 0. The graph of f is a surface with equation z = f (x, y). Let S
be the solid that lies above R and under the graph of f , that is
S = {(x, y, z) ∈ R3 : 0 ≤ z ≤ f (x, y), (x, y) ∈ R}.
Our goal is to find the volume of S.
Step 1: Partition We first take a partition of [a, b]:
a = x0 < x1 < x2 < · · · < xm = b,
where xi = a + i∆x and ∆x = (b − a)/m; and a partition of [c, d]:
c = y0 < y1 < y2 < · · · < yn = d,
where yj = c + j∆y and ∆y = (d − c)/n. This actually divides the rectangle into mn subrectangles:
Rij = [xi−1 × xi ] × [yj−1 , yj ] = {(x, y) : xi−1 ≤ x ≤ xi , yj−1 ≤ y ≤ yj },
each with area ∆A = ∆x∆y.
∗
Step 2: Sample Points In each sub-rectangle Rij we choose a sample point (x∗ij , yij
). Thus
we can approximate the part of S that lies above each Rij by a thin rectangular box with base
∗
Rij and height f (x∗ij , yij
) as shown in figure 1. Then volume of this box is the height of the box
∗
times the area of the base rectangle f (x∗ij , yij
)∆A.
Step 3: Take a Sum If we follow this procedure for all the rectangles and add the volumes
of the corresponding boxes, we get an approximation to the total volume of S:
V ≈
m X
n
X
∗
f (x∗ij , yij
)∆A.
i=1 j=1
This is called a double Riemann sum.
1
z
z=f(x,y)
Height: f (x*
ij,y*
ij)
y
x
Rij
(x*
ij,y*
ij)
Figure 1: Volume and Double Integrals
Step 4: Take a Limit Finally we take a limit and obtain the exact value of the volume
V =
lim
m,n→∞
m X
n
X
∗
f (x∗ij , yij
)∆A.
i=1 j=1
Definition 1. The double integral of f over the rectangle R is
ZZ
m X
n
X
∗
f (x, y) dA = lim
f (x∗ij , yij
)dA
m,n→∞
R
i=1 j=1
if this limit exists. If f (x, y) ≥ 0, then the double integral above represents the volume V of the
solid that lies above the rectangle R and below the surface z = f (x, y).
Example 2. Estimate the volume of the solid that lies above the square R = [0, 2] × [0, 2] and
below the elliptic paraboloid z = 16 − x2 − 2y 2 . Divide R into four equal squares and choose the
sample point to be the upper right corner of each square Rij . Find the double Riemann sum.
Solution
By the basic formula we have
V ≈
2 X
2
X
∗
f (x∗ij , yij
)dA
i=1 j=1
= f (1, 1)∆A + f (1, 2)∆A + f (2, 1)∆A + f (2, 2)∆A
= 13 × 1 + 7 × 1 + 10 × 1 + 4 × 1 = 34.
2
y
(1,2)
R12
R22
(1,1)
R11
(2,2)
(2,1)
R21
x
Figure 2: The Partition of the Square
Example 3. If R = {(x, y) : −1 ≤ x ≤ 1, −2 ≤ y ≤ 2}, evaluate the integral
ZZ p
1 − x2 dA.
R
z=(1-x2)1/2
y=-2
x=-1
x=1
y=2
Figure 3: The Solid in Example 3: A half circular cylinder
Solution This double integral represents the volume of a half circular cylinder. The volume is
the area of a half-disk with radius 1 times the height of the cylinder. Thus
ZZ p
1
1 − x2 dA = π(1)2 × 4 = 2π.
2
R
3
Proposition 4 (Midpoint Rule for Double Integrals).
ZZ
f (x, y) dA ≈
R
m X
n
X
f (x̄i , ȳj )∆A,
i=1 j=1
where x̄i is the midpoint of [xi−1 , xi ] and ȳj is the midpoint of [yj−1 , yj ].
Average Value We define the average value of a function f of two variables defined on a
rectangle R to be
ZZ
1
f (x, y) dA,
fave =
A(R)
R
where A(R) is the area of R.
Properties of Double Integrals Since the double integral is defined as the limit of Riemann
sums, we have
ZZ
ZZ
ZZ
[f (x, y) + g(x, y)] dA =
f (x, y) dA +
g(x, y) dA;
R
R
R
ZZ
ZZ
cf (x, y) dA = c
f (x, y) dA; c ∈ R.
R
R
ZZ
ZZ
If f (x, y) ≥ g(x, y) for all (x, y) in R, then
f (x, y) dA ≥
R
4
g(x, y) dA.
R